Critical points, Hessians and boundary checks

Try

Consider f(x,y)=(x−1)²+2(y+2)²+3. Its smallest value is visibly 3, reached at (1,−2). Both squared terms are nonnegative, so this conclusion is global and needs no numerical search. Now compare g(x,y)=x²−2y². At the origin its first derivatives are zero, but moving along the x-axis increases g while moving along the y-axis decreases it. A zero gradient identifies a stationary candidate, not automatically a minimum.

Predict

Predict what the level curves reveal. Near the first function’s minimum, positive level increments form nested ellipses around one center. For the saddle, positive and negative values occur arbitrarily close to the origin, separated by the zero-level lines x=±√2y. Draw these sign regions before using a classification formula. This sketch answers a mathematical question: can all nearby points lie above, all below, or neither? It should not imply a finite plotted window proves a global extremum.

Experiment

For h(x,y)=2x²+2xy+3y², calculate the gradient (4x+2y,2x+6y). Solving both equations gives only (0,0). Its Hessian, the matrix of second partials, is [[4,2],[2,6]]. The determinant is 20 and the upper-left entry is positive. Complete the square independently: h=2(x+y/2)²+(5/2)y². Both methods establish a strict minimum at zero.

Change the final coefficient from 3 to −1. The Hessian determinant becomes −12, and the completed quadratic has mixed signs. Record what changed in the expression, not merely the classification label.

Understand

For a C² function near an interior critical point, let D=f_xx f_yy−f_xy², evaluated there. If D>0 and f_xx>0, there is a strict local minimum; if D>0 and f_xx<0, a strict local maximum; if D<0, a saddle. If D=0, the test is inconclusive.

The reason is the Taylor expansion f(p+h)=f(p)+(1/2)hᵀHh+o(|h|²). Positive definiteness makes the quadratic term uniformly positive relative to |h|², so a sufficiently small remainder cannot reverse its sign. Negative definiteness reverses the conclusion. Indefiniteness supplies directions of both signs. A zero eigenvalue leaves a direction where higher-order terms can decide the behavior.

Apply

Repair “the Hessian is zero, so the function is flat and has no strict extremum.” The functions x⁴+y⁴, −x⁴−y⁴ and x⁴−y⁴ all have zero Hessian at the origin, yet give a strict minimum, strict maximum and saddle respectively. Higher-order information is essential.

For a continuous function on a closed bounded region, global extrema exist, but interior stationary points are only part of the search. On a rectangle, inspect each edge as a one-variable problem and include its endpoints. For f=x+y on the unit square, the gradient never vanishes, yet the minimum is 0 at (0,0) and maximum 2 at (1,1). Boundary points need not satisfy the interior stationary condition.

Continue

Optional depth connects this lesson to eigenvalues: a real symmetric Hessian has orthogonal principal directions, and the signs of its eigenvalues classify a nondegenerate critical point. This is the same quadratic geometry encountered in linear algebra, now applied to local change.

Write a solution that states the domain before listing candidates. Distinguish local from global conclusions, and explain how boundary cases were exhausted. A computer plot can suggest a candidate but cannot establish that no omitted boundary point is better. Next, constrained optimization will replace separate edge parametrizations with gradients normal to regular constraint surfaces, while retaining these existence and completeness checks.